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Emma's question at Yahoo! Answers regarding solving a trigonometric equation
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Hello Emma,
We are given to solve:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0\) where \(\displaystyle 0^{\circ}\le x\le360^{\circ}\)
If we use the following double-angle identities for sine and cosine:
\(\displaystyle \cos(2\theta)=2\cos^2(\theta)-1\,\therefore\,cos^2(\theta)=\frac{1+\cos(2\theta)}{2}\)
\(\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)\)
then the equation becomes:
\(\displaystyle \frac{1+\cos(2x)}{2}-4\sin(2x)+3=0\)
which we can arrange as:
\(\displaystyle 8\sin(2x)-\cos(x)=7\)
Now, if we define (where \(\displaystyle k\in\mathbb{R},\,0^{\circ}<\alpha<90^{\circ}\)):
\(\displaystyle 8=k\cos(\alpha)\)
\(\displaystyle 1=k\sin(\alpha)\)
then we obtain by division:
\(\displaystyle \tan(\alpha)=\frac{1}{8}\,\therefore\,\alpha=\tan^{-1}\left(\frac{1}{8} \right)\)
and by squaring and adding:
\(\displaystyle 8^2+1^2=65=k^2\left(cos^2(\alpha)+\sin^2(\alpha) \right)=k^2\,\therefore\,k=\sqrt{65}\)
Hence, our equation becomes:
\(\displaystyle \cos(\alpha)\sin(2x)-\sin(\alpha)\cos(x)=\frac{7}{k}\)
Using the angle-difference identity for sine, and substituting for $\alpha$ and $k$ we obtain:
\(\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right) \right)=\frac{7}{\sqrt{65}}\)
Because of the periodicity of the sine function, we may write, where \(\displaystyle k\in\mathbb{Z}\)
(1) \(\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ} \right)=\frac{7}{\sqrt{65}}\)
Now, combining this with the identity \(\displaystyle \sin\left(180^{\circ}-\theta \right)=\sin(\theta)\) we also have:
(2) \(\displaystyle \sin\left(\tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ} \right)=\frac{7}{\sqrt{65}}\)
Taking the inverse sine of both sides of (1), we find:
\(\displaystyle 2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)\)
Solving for $x$, we find:
\(\displaystyle x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)-k\cdot180^{\circ}\)
Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:
\(\displaystyle k=0\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right) \approx33.69006752598^{\circ}\)
\(\displaystyle k=-1\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)+180^{\circ} \approx213.69006752597983^{\circ}\)
Taking the inverse sine of both sides of (2), we find:
\(\displaystyle \tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)\)
Solving for $x$, we find:
\(\displaystyle x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+(2k+1)\cdot90^{\circ}\)
Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:
\(\displaystyle k=0\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+90^{\circ} \approx63.43494882292201^{\circ}\)
\(\displaystyle k=1\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+270^{\circ} \approx243.43494882292202^{\circ}\)
We are given to solve:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0\) where \(\displaystyle 0^{\circ}\le x\le360^{\circ}\)
If we use the following double-angle identities for sine and cosine:
\(\displaystyle \cos(2\theta)=2\cos^2(\theta)-1\,\therefore\,cos^2(\theta)=\frac{1+\cos(2\theta)}{2}\)
\(\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)\)
then the equation becomes:
\(\displaystyle \frac{1+\cos(2x)}{2}-4\sin(2x)+3=0\)
which we can arrange as:
\(\displaystyle 8\sin(2x)-\cos(x)=7\)
Now, if we define (where \(\displaystyle k\in\mathbb{R},\,0^{\circ}<\alpha<90^{\circ}\)):
\(\displaystyle 8=k\cos(\alpha)\)
\(\displaystyle 1=k\sin(\alpha)\)
then we obtain by division:
\(\displaystyle \tan(\alpha)=\frac{1}{8}\,\therefore\,\alpha=\tan^{-1}\left(\frac{1}{8} \right)\)
and by squaring and adding:
\(\displaystyle 8^2+1^2=65=k^2\left(cos^2(\alpha)+\sin^2(\alpha) \right)=k^2\,\therefore\,k=\sqrt{65}\)
Hence, our equation becomes:
\(\displaystyle \cos(\alpha)\sin(2x)-\sin(\alpha)\cos(x)=\frac{7}{k}\)
Using the angle-difference identity for sine, and substituting for $\alpha$ and $k$ we obtain:
\(\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right) \right)=\frac{7}{\sqrt{65}}\)
Because of the periodicity of the sine function, we may write, where \(\displaystyle k\in\mathbb{Z}\)
(1) \(\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ} \right)=\frac{7}{\sqrt{65}}\)
Now, combining this with the identity \(\displaystyle \sin\left(180^{\circ}-\theta \right)=\sin(\theta)\) we also have:
(2) \(\displaystyle \sin\left(\tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ} \right)=\frac{7}{\sqrt{65}}\)
Taking the inverse sine of both sides of (1), we find:
\(\displaystyle 2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)\)
Solving for $x$, we find:
\(\displaystyle x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)-k\cdot180^{\circ}\)
Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:
\(\displaystyle k=0\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right) \approx33.69006752598^{\circ}\)
\(\displaystyle k=-1\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)+180^{\circ} \approx213.69006752597983^{\circ}\)
Taking the inverse sine of both sides of (2), we find:
\(\displaystyle \tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)\)
Solving for $x$, we find:
\(\displaystyle x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+(2k+1)\cdot90^{\circ}\)
Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:
\(\displaystyle k=0\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+90^{\circ} \approx63.43494882292201^{\circ}\)
\(\displaystyle k=1\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+270^{\circ} \approx243.43494882292202^{\circ}\)
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While I was busy composing the above posts, someone else at Yahoo! Answers replied with a much simpler method, which I will outline for the benefit of our members.
We are given to solve:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0\) where \(\displaystyle 0^{\circ}\le x\le360^{\circ}\)
Using a Pythagorean identity, we may write the equation as:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3\left(\sin^2(x)+\cos^2(x) \right)=0\)
We may arrange this as:
\(\displaystyle 4\cos^2(x)-8\sin(x)\cos(x)+3\sin^2(x)=0\)
Factoring, we obtain:
\(\displaystyle \left(2\cos(x)-\sin(x) \right)\left(2\cos(x)-3\sin(x) \right)=0\)
From the first factor, we obtain the solutions (where \(\displaystyle k\in\mathbb{k}\)):
\(\displaystyle x=\tan^{-1}(2)+k\cdot180^{\circ}\)
and for appropriate values of $k$, we obtain:
\(\displaystyle k=0\implies x=\tan^{-1}(2)\approx63.43494882292201^{\circ}\)
\(\displaystyle k=1\implies x=\tan^{-1}(2)+180^{\circ}\approx243.43494882292202^{\circ}\)
From the second factor, we obtain the solutions:
\(\displaystyle x=\tan^{-1}\left(\frac{2}{3} \right)+k\cdot180^{\circ}\)
and for appropriate values of $k$, we obtain:
\(\displaystyle k=0\implies x=\tan^{-1}\left(\frac{2}{3} \right)\approx33.69006752598^{\circ}\)
\(\displaystyle k=1\implies x=\tan^{-1}\left(\frac{2}{3} \right)+180^{\circ}\approx213.69006752597977^{ \circ}\)
I sure wish I had realized this method first!
We are given to solve:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0\) where \(\displaystyle 0^{\circ}\le x\le360^{\circ}\)
Using a Pythagorean identity, we may write the equation as:
\(\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3\left(\sin^2(x)+\cos^2(x) \right)=0\)
We may arrange this as:
\(\displaystyle 4\cos^2(x)-8\sin(x)\cos(x)+3\sin^2(x)=0\)
Factoring, we obtain:
\(\displaystyle \left(2\cos(x)-\sin(x) \right)\left(2\cos(x)-3\sin(x) \right)=0\)
From the first factor, we obtain the solutions (where \(\displaystyle k\in\mathbb{k}\)):
\(\displaystyle x=\tan^{-1}(2)+k\cdot180^{\circ}\)
and for appropriate values of $k$, we obtain:
\(\displaystyle k=0\implies x=\tan^{-1}(2)\approx63.43494882292201^{\circ}\)
\(\displaystyle k=1\implies x=\tan^{-1}(2)+180^{\circ}\approx243.43494882292202^{\circ}\)
From the second factor, we obtain the solutions:
\(\displaystyle x=\tan^{-1}\left(\frac{2}{3} \right)+k\cdot180^{\circ}\)
and for appropriate values of $k$, we obtain:
\(\displaystyle k=0\implies x=\tan^{-1}\left(\frac{2}{3} \right)\approx33.69006752598^{\circ}\)
\(\displaystyle k=1\implies x=\tan^{-1}\left(\frac{2}{3} \right)+180^{\circ}\approx213.69006752597977^{ \circ}\)
I sure wish I had realized this method first!